What is the extraneous solution to these equations? $\dfrac{x^2 - 3}{x + 8} = \dfrac{4x - 7}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 3}{x + 8} (x + 8) = \dfrac{4x - 7}{x + 8} (x + 8)$ $ x^2 - 3 = 4x - 7$ Subtract $4x - 7$ from both sides: $ x^2 - 3 - (4x - 7) = 4x - 7 - (4x - 7)$ $ x^2 - 3 - 4x + 7 = 0$ $ x^2 + 4 - 4x = 0$ Factor the expression: $ (x - 2)(x - 2) = 0$ Therefore $x = 2$ The original expression is defined at $x = 2$ and $x = 2$, so there are no extraneous solutions.